The diagram shows part of the curve $y = \sin x$. A is a point on the curve that lies on the line $x = \frac{\pi}{3}$. The line AC is perpendicular to the x-axis.

1. Find the exact coordinates of A

At $x = \frac{\pi}{3}$,
$y = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Coordinates of A are $\left(\frac{\pi}{3}, \frac{\sqrt{3}}{2}\right)$.

2. Find the gradient of the tangent at A

$y = \sin x$
$\frac{dy}{dx} = \cos x$
At $x = \frac{\pi}{3}$, gradient of tangent $m_t = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

3. Find the gradient of the normal at A

Gradient of normal $m_n = -\frac{1}{m_t} = -\frac{1}{\frac{1}{2}} = -2$.

4. Find the coordinates of B by setting up the equation of the normal

Let the coordinates of B be $(x_B, 0)$.
Using the gradient of the normal:
$\frac{0 - \frac{\sqrt{3}}{2}}{x_B - \frac{\pi}{3}} = -2$
$-\frac{\sqrt{3}}{2} = -2\left(x_B - \frac{\pi}{3}\right)$
$x_B - \frac{\pi}{3} = \frac{\sqrt{3}}{4}$
$x_B = \frac{\pi}{3} + \frac{\sqrt{3}}{4}$
Coordinates of B are $\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}, 0\right)$.